How Long Does It Take for Hot Water in a Tank to Cool Down to 50°C?

[oblomov]

a cylindrical tank of height 2m with a cross sectional area of 0.5m^2 contains hot water at 80oC , 125kPa.it is in a room with temperature To=20oC , so it slowly loses energy to the room air proportional to the temperature difference as Qloss=CA(T-To) with the tank surface area,A, C is a constant.for different values of the constant C,estimate the time it takes to bring the water to 50oC.make enough simlifying assumptions so that you can solve the problem mathematically, that is find a formula for T(t).
T-To=delta T=derivative of T(t) .should i integrate after doing this?but i don t know Qloss.at t=0 Qloss=60CA .

 

[Valkarie]

so i am stuck.please help.The equation that describes the rate of change of temperature of the hot water is dT/dt = -Qloss/(mCp)where m is the mass of the hot water, Cp is the specific heat capacity of the water, and Qloss is the rate of energy loss to the room. To solve the problem, we first need to calculate the mass of the hot water. Since the tank has a cross-sectional area of 0.5 m^2 and a height of 2 m, the volume of the tank is V = 0.5 x 2 = 1 m^3. The density of water is 1000 kg/m^3, so the mass of the hot water is m = V x density = 1000 kg. Next, we need to calculate the specific heat capacity of the water. This can be calculated using the equation Cp = c x mwhere c is the specific heat capacity of water at the given temperature and m is the mass of the water. For water at 80°C, the specific heat capacity is 4.187 kJ/kgK. Thus, the specific heat capacity of the hot water is Cp = 4.187 x 1000 = 4187 kJ/kgK. Now we can solve for the time it takes for the hot water to cool to 50°C. Substituting the values for m, Cp, and Qloss into the equation for dT/dt, we get dT/dt = -60CA/(4187 x 1000) Integrating both sides of the equation with respect to time, we get ∫dT/dt dt = ∫(-60CA/(4187 x 1000)) dt T(t) - T(0) = -60CAt/(4187 x 1000) Substituting the initial temperature (T0 = 80°C) and the final temperature (Tf = 50°C), we get 50 - 80 = -60CAt/(4187 x 1000) Solving for t, we get t = (30 x 4187 x 1000)/(60CA) Thus, the time it takes for the hot water to cool from 80°C to

 

[piercas]

I would approach this thermodynamic question by first making some simplifying assumptions to solve the problem mathematically. These assumptions could include assuming that the tank is well-insulated and that there is no heat transfer through the bottom of the tank. I would also assume that the temperature of the room remains constant at 20oC throughout the entire process.

Using these assumptions, we can use the first law of thermodynamics to write an energy balance equation for the system:

dU/dt = Qin - Qout

Where dU/dt is the rate of change of internal energy, Qin is the heat transfer into the tank, and Qout is the heat transfer out of the tank. We can also assume that the water in the tank is well-mixed, so the temperature is uniform throughout.

Using the equation given in the problem, we can write Qout as:

Qout = CA(T - To)

Substituting this into the energy balance equation and rearranging, we get:

dU/dt + CA(T - To) = 0

We can also assume that the specific heat of water, c, and the density of water, ρ, are constant. This allows us to write the internal energy, U, in terms of the temperature and volume of the water:

U = mcvT

Where m is the mass of water in the tank and V is the volume.

Substituting this into the energy balance equation and simplifying, we get:

d(T)/dt + (cρA/2V)(T - To) = 0

This is a first-order linear differential equation that we can solve using the method of integrating factors. Multiplying both sides by the integrating factor, e^(cρA/2V)t, we get:

e^(cρA/2V)t d(T)/dt + (cρA/2V)e^(cρA/2V)t(T - To) = 0

The left side of this equation can be rewritten as the derivative of the product of e^(cρA/2V)t and T:

d/dt (e^(cρA/2V)t T) = 0

Integrating both sides with respect to t, we get:

e^(cρA/2V)t T = C1

Where C1 is a constant of integration. Solving for T, we get:

T = C1e